Question

If two zeroes of the polynomial$x^4-6x^3-26x^2+138x-35$ are $2\pm \sqrt{3},$find other zeroes.

Answer 1

Solve for the zeroes of the given polynomial

Given $x^4-6x^3-26x^2+138x-35$.

The given equation is a polynomial of degree $4,$hence there will be total $4$roots.

Since $2+\sqrt{3}$and $2-\sqrt{3}$are zeroes of the given polynomial $f\left(x\right)$.

$$\begin{gathered} \therefore (x-(2+\sqrt{3}\left)\right)(x-(2-\sqrt{3}\left)\right)=0 \\ (x-2-\sqrt{3})(x-2+\sqrt{3})=0 \end{gathered}$$

On multiplying the above equation we get,$x^2-4x+1,$ that this is a factor of a given polynomial$f\left(x\right)$.

Now, dividing $f\left(x\right)$by $g\left(x\right),$the quotient will also be a factor of$f\left(x\right)$ and the remainder will be$0$.

$x^2-4x+1x^2-2x-35x^{4}6x^3-26x^2+138x-35x^{4}4x^3+x^2-2x^3-27x^2+138x-35-2x^3+8x^2-2x-35x^2+140x-35-35x^2+140x-350$

So, $x^4-6x^3-26x^2+138x-35$

$=(x^2-4x+1)(x^2–2x-35)$

On factorization $(x^2–2x-35)$we get,

$x^2–(7-5)x-35$

$x^2–7x+5x+35=0$

$$\begin{gathered} x(x-7)+5(x-7)=0 \\ (x+5)(x-7)=0 \end{gathered}$$

So, the zeroes are given by: $x=-5$and$x=7$.

Hence, all four zeroes of the given polynomial equation are$2+\sqrt{3},2-\sqrt{3},-5$ and$7$.